Chapter 8 Solutions - Soln0861

Transcript

Chapter 8, Solution 61. For t > 0, we obtain the natural response by considering the circuit below. a 1H iL + 4 vC 0.25F 6  At node a, v C /4 + 0.25dv C /dt + i L = 0 (1) But, v C = 1di L /dt + 6i L (2) Combining (1) and (2), (1/4)di L /dt + (6/4)i L + 0.25d2i L /dt2 + (6/4)di L /dt + i L = 0 d2i L /dt2 + 7di L /dt + 10i L = 0 s2 + 7s + 10 = 0 = (s + 2)(s + 5) or s 1,2 = -2, -5 Thus, i L (t) = i L () + [Ae-2t + Be-5t], where i L () represents the final inductor current = 4(4)/(4 + 6) = 1.6 i L (t) = 1.6 + [Ae-2t + Be-5t] and i L (0) = 1.6 + [A+B] or -1.6 = A+B (3) di L /dt = [-2Ae-2t - 5Be-5t] and di L (0)/dt = 0 = -2A – 5B or A = -2.5B (4) From (3) and (4), A = -8/3 and B = 16/15 i L (t) = 1.6 + [-(8/3)e-2t + (16/15)e-5t] v(t) = 6i L (t) = {9.6 + [-16e-2t + 6.4e-5t]} V v C = 1di L /dt + 6i L = [ (16/3)e-2t - (16/3)e-5t] + {9.6 + [-16e-2t + 6.4e-5t]} v C = {9.6 + [-(32/3)e-2t + 1.0667e-5t]} i(t) = v C /4 = {2.4 + [-2.667e-2t + 0.2667e-5t]} A